All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
The Triangle's Bottom (Posted on 2008-02-05) Difficulty: 2 of 5
Let T be the set of triangular numbers and T* be the set of all products of any two triangular numbers. Show that:

1. Among elements of T, each of the digits 0,1,5 and 6 occur in the units place twice as frequently as each of the digits 3 and 8. (More precisely, if MBk is the set of elements of T that are less than B and end in k, then, e.g., MB1/MB8 approaches 2 as B approaches infinity.)

2. None of the elements of T* end in 2 or 7.

See The Solution Submitted by FrankM    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution | Comment 1 of 4

Each triangular number T(i) equals the previous triangular number plus i:  T(i) = T(i-1)+i.

So:


 i   T(i)
 1     1
 2     3
 3     6
 4    10
 5    15
 6    21
 7    28
 8    36
 9    45
10    55
11    66
12    78
13    91
14   105
15   120
16   136
17   153
18   171
19   190
20   210
21   231
22   253
23   276
24   300
25   325
26   351
27   378
28   406
29   435
30   465
31   496
32   528
33   561
34   595
35   630
36   666
37   703
38   741
39   780
40   820

Note that there is a cycle of 20 in the last digits. For i itself, there's a cycle of 10 in the last digits, but it is not until i=21 that T(i) also returns to a last digit of 1, as at T(1). From then on, due to the nature of addition, the length-20 cycle repeats.

In the cycle of 20, there are the following counts of last digits:

final
digit  count
 0      4
 1      4
 2      0
 3      2
 4      0
 5      4
 6      4
 7      0
 8      2
 9      0
 


Therefore, over the long run, the occurrence of each of 0, 1, 5 and 6 is twice as frequent as either of 3 or 8.

When taking the products of these numbers, which never end in 2, 4, 7 or 9, if either one ends in a zero, the product will end in a zero. If either ends in a 1, the product will end in the same digit as the other triangular number and never be 2, 4, 7 or 9.  The other possible pairs of ending digits result in products ending as follows:

1  3  5  6  8
3  9  5  8  4
5  5  5  0  0
6  8  0  6  8
8  4  0  8  4

None of these is a 2 or a 7.


  Posted by Charlie on 2008-02-05 14:48:22
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (9)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information